Artificial gravity meeting zero gravity

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In a hypothetical spaceship with two concentric hoops, one rotating to create artificial gravity, an occupant transitioning from the moving hoop to the stationary one would initially experience weightlessness until they make contact with a surface that imparts tangential motion. This contact is essential for experiencing artificial gravity, as inertia from the moving hoop is what provides the sensation of weight. If the occupant accelerates in the opposite direction of the hoop's rotation, they would feel lighter, while moving in the same direction would make them feel heavier. The dynamics of lift and weight in such a rotating environment are complex, as a plane flying within this space station would need to account for the rotational speed and air movement to maintain stability. Ultimately, the interplay of these forces could allow for unique flight experiences, including the possibility of muscle-powered flight in a low-gravity setting.
  • #31
Funny that these same calculations result in a counter-proof to the hollow-Earthers (Koreshian cosmogony):
If we did inhabit the inside surface of a hollow sphere the size of the Earth, the centripetal acceleration at the equator would be over 3G...
 
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  • #32
dbell5 said:
Funny that these same calculations result in a counter-proof to the hollow-Earthers (Koreshian cosmogony):
If we did inhabit the inside surface of a hollow sphere the size of the Earth, the centripetal acceleration at the equator would be over 3G...

So there is more force than just gravity holding the planet together?
 
  • #33
dbell5 said:
Funny that these same calculations result in a counter-proof to the hollow-Earthers (Koreshian cosmogony):
If we did inhabit the inside surface of a hollow sphere the size of the Earth, the centripetal acceleration at the equator would be over 3G...

How do you figure 3Gs?

If that were true, it would apply to the outer surface as much as the inner surface. What you are suggesting is that, standing on the outer surface of the Earth, I am experiencing a 3G pull away from the Earth!
 
  • #34
DaveC426913 said:
How do you figure 3Gs?

If that were true, it would apply to the outer surface as much as the inner surface. What you are suggesting is that, standing on the outer surface of the Earth, I am experiencing a 3G pull away from the Earth!

Huh! That would appear to be so! (Actually, 2G away, since the Earth's field is 1G inwards at the surface, vs. 0G inside a hollow shell.)

OK, here's how I got there:
a = v^2/r
Circumference of Earth is about 41,000 km, or 41e6 meters
Rotation is once in 86,400 seconds, for a tangential velocity of about 474 m/s.
Radius is about 6,500 km, which is where I went wrong - I divided by 6,500, not 6.5e6.
I got 34 m/s^2, should be 0.034 m/s^2

OK, so we're in no immediate danger of flying off into space!

Dave
 
  • #35
A corollary to the above that I never thought of is that, even if the Earth were a perfect, homogeneous sphere, and ignoring the bulky clothes, I'd weigh nearly 5 pounds more at the poles than I do here at 37°N!
 
  • #36
dbell5 said:
A corollary to the above that I never thought of is that, even if the Earth were a perfect, homogeneous sphere, and ignoring the bulky clothes, I'd weigh nearly 5 pounds more at the poles than I do here at 37°N!

So that's why Santa is so heavy.
 

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